Affine varieties, as the common zeroes of polynomials over some field, are irreducible algebraic sets and they're relatively easy to understand when polynomials are over $\Bbb C$. However, how can I determine that an algebraic set is a variety (i.e. irreducible) over a finite field $\Bbb F_q$, where q is a power of a prime. And more importantly, how to find its dimension?

For instance, in order to make everything easy, let $f=x^2+3$ and our finite field be $\Bbb F_7$. The algebraic set defined by $f$ over $\Bbb F_7$ is the set $$V= \{y\in acl(\Bbb F_7) : f(y)=0 \}$$

Therefore $2,5,9,12 \in V$, as $2^2+3 \equiv5^2+3 \equiv 9^2+3 \equiv 0 $ (mod $7$), and there are infinitely such elements in $V$. Equivalently, $V$ consists of the numbers $2$ mod $7$ and $5$ mod $7$. So, I think that $V$ is defined by union of smaller algebraic sets $x-2$ and $x-5$. So, I think that V is not a variety. But, are these smaller components of $V$ varieties? If they're so, what are their dimension? More generally, how can I find dimension of a variety defined over a finite field. For example, what is dimension of the variety defined by $x-2$ over $\Bbb F_7$ ?

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    $\begingroup$ 2,9 and 5,12 are the same elements modulo 7. $\endgroup$ – Wuestenfux Aug 14 '19 at 18:03
  • $\begingroup$ @Wuestenfux I mean 9, 12 as elements of algebraic closure of F_7, which is union of all finite fields with 7^n elements for each n. So, 9 as itself stays in it, i guess? $\endgroup$ – offret Aug 14 '19 at 18:23
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    $\begingroup$ In any field of characteristic $7$ you have $2=9$ and $5=12$. In particular for any algebraic closure of $\Bbb{F}_7$. $\endgroup$ – Servaes Aug 14 '19 at 19:21
  • $\begingroup$ Chiming in with others. For a different way of looking at it: You do know that $\Bbb{F}_{7^n}\cong\Bbb{F}_7[x]/\langle p(x)\rangle$ for any irreducible polynomial $p(x)$ of degree from $\Bbb{F}_7[x]$. In the ring $\Bbb{F}_7[x]$ we have $2=9$, so the same holds for all the quotients. $\endgroup$ – Jyrki Lahtonen Aug 14 '19 at 19:50
  • $\begingroup$ And yet another way of looking at it: The algebraic closure is an extension of $\Bbb{F}_7$. Meaning that the elements that are equal in $\Bbb{F}_7$ remain equal in all the extensions. For example $2/1$ and $4/2$ are the same elements in $\Bbb{Q}$, so they remain the same elements in $\Bbb{R}$ and $\Bbb{C}$. You need to review finite fields and field extension sooner rather than later. $\endgroup$ – Jyrki Lahtonen Aug 14 '19 at 19:58

Your $V$ consists of the union of two zero-dimensional varieties, namely the points $y = 2$ and $y = -2$. There are many ways to define the dimension of a variety, such as the Krull dimension (in the affine case) or the transcendence degree of the function field.

You can determine whether an affine variety is irreducible by checking whether its ring of functions is an integral domain.


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